computing the stanley depth

Journal of Algebra 323 (2010) 2943–2959

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Journal of Algebra

www.elsevier.com/locate/jalgebra

Computing the Stanley depth ✩

Dorin Popescu a,∗, Muhammad Imran Qureshi b

a Institute of Mathematics “Simion Stoilow”, University of Bucharest, PO Box 1-764, Bucharest 014700, Romaniab Abdus Salam School of Mathematical Sciences, GC University, Lahore, 68-B New Muslim town Lahore, Pakistan

a r t i c l e i n f o a b s t r a c t

Article history:Received 25 November 2009Available online 16 December 2009Communicated by Steven Dale Cutkosky

Keywords:Monomial idealsStanley decompositionsStanley depth

Let Q and Q ′ be two monomial primary ideals of a polynomialalgebra S over a field. We give an upper bound for the Stanleydepth of S/(Q ∩ Q ′) which is reached if Q , Q ′ are irreducible.Also we show that Stanley’s Conjecture holds for Q 1 ∩ Q 2,S/(Q 1 ∩ Q 2 ∩ Q 3), (Q i)i being some irreducible monomial idealsof S .

© 2009 Elsevier Inc. All rights reserved.

Introduction

Let K be a field and S = K [x1, . . . , xn] be the polynomial ring over K in n variables and M a finitelygenerated multigraded (i.e. Z

n-graded) S-module. Given z ∈ M a homogeneous element in M andZ ⊆ {x1, . . . , xn}, let zK [Z ] ⊂ M be the linear K -subspace of all elements of the form zf , f ∈ K [Z ].This subspace is called Stanley space of dimension |Z |, if zK [Z ] is a free K [Z ]-module. A Stanleydecomposition of M is a presentation of the K -vector space M as a finite direct sum of Stanleyspaces D: M = ⊕r

i=1 zi K [Zi]. Set sdepth D = min{|Zi |: i = 1, . . . , r}. The number

sdepth(M) := max{

sdepth(D): D is a Stanley decomposition of M}

is called the Stanley depth of M . This is a combinatorial invariant which has some common propertieswith the homological invariant depth. Stanley conjectured (see [17]) that sdepth M � depth M , but thisconjecture is still open for a long time in spite of some results obtained mainly for n � 5 (see [1,16,8,

✩ The authors would like to express their gratitude to ASSMS of GC University Lahore for creating a very appropriateatmosphere for research work. This research is partially supported by HEC Pakistan. The first author was mainly supportedby CNCSIS Grant ID-PCE no. 51/2007.

* Corresponding author.E-mail addresses: [email protected] (D. Popescu), [email protected] (M.I. Qureshi).

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2944 D. Popescu, M.I. Qureshi / Journal of Algebra 323 (2010) 2943–2959

2,12,13]). An algorithm to compute the Stanley depth is given in [9] and was used here to find severalexamples. Very important in our computations were the results from [3,6,15].

Let Q , Q ′ be two monomial primary ideals such that dim S/(Q + Q ′) = 0. Then

sdepth S/(

Q ∩ Q ′) � max

{min

{dim

(S/Q ′),⌈dim(S/Q )

2

⌉},min

{dim(S/Q ),

⌈dim(S/Q ′)

2

⌉}},

and the bound is reached when Q , Q ′ are non-zero irreducible monomial ideals (see Proposition 2.2,or more general in Corollary 2.4), � a

2 being the smallest integer � a/2, a ∈ Q.Let Q 1, Q 2, Q 3 be three non-zero irreducible monomial ideals of S . If dim S/(Q 1 + Q 2) = 0 then

sdepth(Q 1 ∩ Q 2) �⌈

dim(S/Q 1)

2

⌉+

⌈dim(S/Q 2)

2

⌉

(see Lemma 4.3, or more general in Theorem 4.5). In this case, our bound is better than the boundgiven by [10] and [11] (see Remark 4.2). Using these results we show that sdepth(Q 1 ∩ Q 2) �depth(Q 1 ∩ Q 2), and

sdepth S/(Q 1 ∩ Q 2 ∩ Q 3) � depth S/(Q 1 ∩ Q 2 ∩ Q 3),

that is Stanley’s Conjecture holds for Q 1 ∩ Q 2 and S/(Q 1 ∩ Q 2 ∩ Q 3) (see Theorems 5.6, 5.9).

1. A lower bound for Stanley’s depth of some cycle modules

We start with few simple lemmas which we include for the completeness of our paper.

Lemma 1.1. Let Q be a monomial primary ideal in S = K [x1, . . . , xn]. Suppose that√

Q = (x1, . . . , xr) where1 � r � n, Then there exists a Stanley decomposition

S/Q =⊕

uK [xr+1, . . . , xn],

where the sum runs on monomials u ∈ K [x1, . . . , xr] \ (Q ∩ K [x1, . . . , xr]).

Proof. Given u, v ∈ K [x1, . . . , xr] \ (Q ∩ K [x1, . . . , xr]) and h, g ∈ K [xr+1, . . . , xn] with uh = vg then weget u = v , g = h. Thus the given sum is direct. Note that there exist just a finite number of monomialsin K [x1, . . . , xr] \ (Q ∩ K [x1, . . . , xr]). Let 0 �= α ∈ (S \ Q ) be a monomial. Then α = u f , where f ∈K [xr+1, . . . , xn] and u ∈ K [x1, . . . , xr]. Since α /∈ Q we have u /∈ Q . Thus S/Q ⊂ ⊕

uK [xr+1, . . . , xn],the other inclusion being trivial. �Lemma 1.2. Let Q be a monomial primary ideal in S = K [x1, . . . , xn]. Then sdepth S/Q = dim S/Q =depth S/Q .

Proof. Let dim S/Q = n − r for some 0 � r � n. We have dim S/Q � sdepth S/Q by [1, Theorem 2.4].Renumbering variables we may suppose that

√Q = (x1, . . . , xr). Using the above lemma we get the

converse inequality. As S/Q is Cohen Macaulay it follows dim S/Q = depth S/Q , which is enough. �Lemma 1.3. Let I , J be two monomial ideals of S = K [x1, . . . , xn]. Then

sdepth(

S/(I ∩ J ))� max

{min

{sdepth(S/I), sdepth

(I/(I ∩ J )

)},

min{

sdepth(S/ J ), sdepth(

J/(I ∩ J ))}}

.

D. Popescu, M.I. Qureshi / Journal of Algebra 323 (2010) 2943–2959 2945

Proof. Consider the following exact sequence of S-modules:

0 → I/(I ∩ J ) → S/(I ∩ J ) → S/I → 0.

By [14, Lemma 2.2], we have

sdepth(

S/(I ∩ J ))� min

{sdepth(S/I), sdepth

(I/(I ∩ J )

)}. (1)

Similarly, we get

sdepth(

S/(I ∩ J ))� min

{sdepth(S/ J ), sdepth

(J/(I ∩ J )

)}. (2)

The proof ends using (1) and (2). �Proposition 1.4. Let Q , Q ′ be two monomial primary ideals in S = K [x1, . . . , xn] with different associatedprime ideals. Suppose that

√Q = (x1, . . . , xt),

√Q ′ = (xr+1, . . . , xn) for some integers t, r with 0 � r � t � n.

Then

sdepth(

S/(

Q ∩ Q ′))� max

{min

v

{r, sdepth

(Q ′ ∩ K [xt+1, . . . , xn]), sdepth

((Q ′ : v

) ∩ K [xt+1, . . . , xn])}

,

minw

{n − t, sdepth

(Q ∩ K [x1, . . . , xr]

), sdepth

((Q : w) ∩ K [x1, . . . , xr]

)}},

where v, w run in the set of monomials containing only variables from {xr+1, . . . , xt}, w /∈ Q , v /∈ Q ′ .

Proof. If Q , or Q ′ is zero then the inequality holds trivially. If r = 0 then Q ∩ K [x1, . . . , xr] =Q ∩ K = 0, and the inequality is clear. A similar case is t = n. Thus we may suppose 1 � r � t < n.Applying Lemma 1.3 it is enough to show that

sdepth(

Q ′/(

Q ∩ Q ′)) � min{

sdepth(

Q ′ ∩ K [xt+1, . . . xn]), sdepth

((Q ′ : v

) ∩ K [xt+1, . . . xn])}

,

where v is a monomial of K [xr+1, . . . , xn] \ (Q ∩ Q ′). We have a canonical injective map

Q ′/(

Q ∩ Q ′) → S/Q .

By Lemma 1.1 we get

Q ′/(

Q ∩ Q ′) = Q ′ ∩(⊕

uK [xt+1, . . . , xn])

=⊕(

Q ′ ∩ uK [xt+1, . . . , xn]),

where u runs in the monomials of K [x1, . . . , xt] \ Q . Here

Q ′ ∩ uK [xt+1, . . . , xn] = u(

Q ′ ∩ K [xt+1, . . . , xn])

if u ∈ K [x1, . . . , xr]

and

Q ′ ∩ uK [xt+1, . . . , xn] = u((

Q ′ : u) ∩ K [xt+1, . . . , xn]) if u /∈ K [x1, . . . , xr].


If u ∈ Q ′ then Q ′ : u = S . We have

Q ′/(

Q ∩ Q ′) =(⊕

u(

Q ′ ∩ K [xt+1, . . . , xn])) ⊕(⊕

zK [xt+1, . . . , xn])

⊕(⊕

uv((

Q ′ : v) ∩ K [xt+1, . . . , xn])),

where the sum runs for all monomials u ∈ (K [x1, . . . , xr] \ Q ), z ∈ Q ′ \ Q and v ∈ K [xr+1, . . . , xt],v /∈ Q ′ ∪ Q . Now it is enough to apply [14, Lemma 2.2] to get the above inequality. �Theorem 1.5. Let Q and Q ′ be two irreducible monomial ideals of S. Then

sdepthS S/(

Q ∩ Q ′) � max

{min

{dim

(S/Q ′),⌈dim(S/Q ) + dim(S/(Q + Q ′))

2

⌉},

min

{dim(S/Q ),

⌈dim(S/Q ′) + dim(S/(Q + Q ′))

2

⌉}}.

Proof. If the associated prime ideals of Q , Q ′ are the same then the above inequality says thatsdepthS S/(Q ∩ Q ′) � dim S/Q , which follows from Lemma 1.2. Thus we may suppose that theassociated prime ideals of Q , Q ′ are different. We may suppose that Q is generated in variables{x1, . . . , xt} and Q ′ is generated in variables {xr+1, . . . , xp} for some integers 0 � r � t � p � n. Sincedim(S/Q ) = n − t , dim(S/Q ′) = n − p + r and dim(S/(Q + Q ′)) = n − p we get

n − t −⌊

p − t

2

⌋=

⌈(n − t) + (n − p)

2

⌉=

⌈dim(S/Q ) + dim(S/(Q + Q ′))

2

⌉,

� a2 � being the biggest integer � a/2, a ∈ Q. Similarly, we have

n − p + r −⌊

r

2

⌋=

⌈dim(S/Q ′) + dim(S/(Q + Q ′))

2

⌉.

On the other hand by [6], and [15, Theorem 2.4] sdepth(Q ′ ∩ K [xt+1, . . . , xn]) = n − t − � p−t2 � and

sdepth(Q ∩ K [x1, . . . , xr, xp+1, . . . , xn]) = n − p + r − � r2 �. In fact, the quoted result says in particular

that sdepth of each irreducible ideal L depends only on the number of variables of the ring andthe number of variables generating L (a description of irreducible monomial ideals is given in [18]).Since (Q ′ : v) ∩ K [xt+1, . . . , xn] is still an irreducible ideal generated by the same variables as Q ′ weconclude that

sdepth((

Q ′ : v) ∩ K [xt+1, . . . , xn]) = sdepth

(Q ′ ∩ K [xt+1, . . . , xn]),

v /∈ Q ′ being any monomial. Similarly,

sdepth((Q : w) ∩ K [x1, . . . , xr, xp+1, . . . , xn]) = sdepth

(Q ∩ K [x1, . . . , xr, xp+1, . . . , xn]

).

It follows that our inequality holds if p = n by Proposition 1.4.Set S ′ = K [x1, . . . , xp], q = Q ∩ S ′ , q′ = Q ′ ∩ S ′ . As above (case p = n) we get

sdepthS ′ S ′/(q ∩ q′) � max

{min

{dim

(S ′/q′),⌈dim(S ′/q)

2

⌉},min

{dim

(S ′/q

),

⌈dim(S ′/q′)

2

⌉}}

= max

{min

{r,

⌈p − t

2

⌉},min

{p − t,

⌈r

2

⌉}}.


Using [9, Lemma 3.6], we have

sdepthS

(S/

(Q ∩ Q ′)) = sdepthS

(S/

(q ∩ q′)S

) = n − p + sdepthS ′(

S ′/(q ∩ q′)).

It follows that

sdepthS

(S/

(Q ∩ Q ′)) � n − p + max

{min

{r,

⌈p − t

2

⌉},min

{p − t,

⌈r

2

⌉}}

= max

{min

{n − p + r,n − p +

⌈p − t

2

⌉},min

{n − t,n − p +

⌈r

2

⌉}}

= max

{min

{n − p + r,n − t −

⌊p − t

2

⌋},min

{n − t,n − p + r −

⌊r

2

⌋}},

which is enough. �2. An upper bound for Stanley’s depth of some cycle modules

Let Q , Q ′ be two monomial primary ideals of S . Suppose that Q is generated in variables{x1, . . . , xt} and Q ′ is generated in variables {xr+1, . . . , xn} for some integers 1 � r � t < n. Thus theprime ideals associated to Q ∩ Q ′ have dimension � 1 and it follows depth(S/(Q ∩ Q ′)) � 1. Thensdepth(S/(Q ∩ Q ′)) � 1 by [5, Corollary 1.6], or [7, Theorem 1.4]. Let D: S/(Q ∩ Q ′) = ⊕s

i=1 ui K [Zi]be a Stanley decomposition of S/(Q ∩ Q ′) with sdepth D = sdepth(S/(Q ∩ Q ′)). Thus |Zi | � 1 forall i. Renumbering (ui, Zi) we may suppose that 1 ∈ u1 K [Z1], so u1 = 1. Note that Zi cannot havemixed variables from {x1, . . . , xr} and {xt+1, . . . , xn} because otherwise ui K [Zi] will be not a freeK [Zi]-module. As | Z1 |� 1 we may have either Z1 ⊂ {x1, . . . , xr} or Z1 ⊂ {xt+1, . . . , xn}.

Lemma 2.1. Suppose Z1 ⊂ {x1, . . . , xr}. Then sdepth(D) � min{r, �n−t2 }.

Proof. Clearly sdepth(D) �| Z1 |� r. Let a ∈ N be such that xai ∈ Q ′ for all t < i � n. Let T =

K [yt+1, . . . , yn] and ϕ : T → S be the K -morphism given by yi → xai . The composition map ψ : T →

S → S/(Q ∩ Q ′) is injective. Note also that we may consider Q ′ ∩ K [xt+1, . . . , xn] ⊂ S/(Q ∩ Q ′) sinceQ ∩ K [xt+1, . . . , xn] = 0. We have

(yt+1, . . . , yn) = ψ−1(Q ′ ∩ K [xt+1, . . . , xn]) =⊕

ψ−1(u j K [Z j] ∩ Q ′ ∩ K [xt+1, . . . , xn]).

If u j K [Z j] ∩ Q ′ ∩ K [xt+1, . . . , xn] �= 0 then u j ∈ K [xt+1, . . . , xn]. Also we have Z j ⊂ {xt+1, . . . , xn},otherwise u j K [Z j] is not free over K [Z j]. Moreover, if ψ−1(u j K [Z j] ∩ Q ′ ∩ K [xt+1, . . . , xn]) �= 0 then

u j = xbt+1t+1 . . . xbn

n , bi ∈ N is such that if xi /∈ Z j , t < i � n, then a | bi , let us say bi = aci for some ci ∈ N.

Denote ci = � bia when xi ∈ Z j . We get

ψ−1(u j K [Z j] ∩ Q ′ ∩ K [xt+1, . . . , xn]) = y

ct+1t+1 . . . ycn

n K [V j],

where V j = {yi: t < i � n, xi ∈ Z j}. Thus ψ−1(u j K [Z j] ∩ Q ′ ∩ K [xt+1, . . . , xn]) is a Stanley spaceof T and so D induces a Stanley decomposition D′ of (yt+1, . . . , yn) such that sdepth(D) �sdepth(D′) � sdepth(yt+1, . . . , yn) because | Z j |=| V j |. Consequently sdepth(D) � �n−t

2 by [3] andso sdepth(D) � min{r, �n−t

2 }.Note also that if t = n, or r = 0 then the same proof works; so sdepth S/(Q ∩ Q ′) = 0, which is

clear because depth S/(Q ∩ Q ′) = 0 (see [5, Corollary 1.6]). �


Proposition 2.2. Let Q , Q ′ be two non-zero monomial primary ideals of S with different associated primeideals. Suppose that dim(S/(Q + Q ′)) = 0. Then

sdepthS

(S/

(Q ∩ Q ′))

� max

{min

{dim

(S/Q ′),⌈dim(S/Q )

2

⌉},min

{dim(S/Q ),

⌈dim(S/Q ′)

2

⌉}}.

Proof. If one of Q , Q ′ is of dimension zero then depth(S/(Q ∩ Q ′)) = 0 and so by [5, Corol-lary 1.6] (or [7, Theorem 1.4]) sdepth(S/(Q ∩ Q ′)) = 0, that is the inequality holds trivially. Thuswe may suppose after renumbering of variables that Q is generated in variables {x1, . . . , xt} andQ ′ is generated in variables {xr+1, . . . , xp} for some integers t, r, p with 1 � r � t < p � n, or0 � r < t � n. By hypothesis we have p = n. Let D be the Stanley decomposition of S/(Q ∩ Q ′)such that sdepth(D) = sdepth(S/(Q ∩ Q ′)). Let Z1 be defined as in Lemma 2.1, that is K [Z1] is theStanley space corresponding to 1. If Z1 ⊂ {x1, . . . , xr} then by Lemma 2.1,

sdepth(D) � min

{r,

⌈n − t

2

⌉}= min

{dim

(S/Q ′),⌈dim(S/Q )

2

⌉}.

If Z1 ⊂ {xt+1, . . . , xn} we get analogously

sdepth(D) � min

{n − t,

⌈r

2

⌉}= min

{dim(S/Q ),

⌈dim(S/Q ′)

2

⌉},

which shows our inequality. �Theorem 2.3. Let Q and Q ′ be two non-zero monomial primary ideals of S with different associated primeideals. Then

sdepthS S/(

Q ∩ Q ′) � max

{min

{dim

(S/Q ′),⌈dim(S/Q ) + dim(S/(Q + Q ′))

2

⌉},

min

{dim(S/Q ),

⌈dim(S/Q ′ + dim(S/(Q + Q ′)))

2

⌉}}.

Proof. As in the proof of Proposition 2.2 we may suppose that Q is generated in variables {x1, . . . , xt}and Q ′ is generated in variables {xr+1, . . . , xp} for some integers 1 � r � t < p � n, or 0 � r < t � nbut now we have not in general p = n. Set S ′ = K [x1, . . . , xp], q = Q ∩ S ′ , q′ = Q ′ ∩ S ′ . Using Proposi-tion 2.2 we get

sdepthS

(S/

(q ∩ q′)) � max

{min

{dim

(S/q′),⌈dim(S/q)

2

⌉},min

{dim(S/q),

⌈dim(S/q′)

2

⌉}}.

By [9, Lemma 3.6] we have

sdepthS

(S/

(Q ∩ Q ′)) = sdepthS

(S/

(q ∩ q′)S

) = n − p + sdepthS ′(

S ′/(q ∩ q′)).

As in the proof of Theorem 1.5, it follows that


sdepthS

(S/

(Q ∩ Q ′)) � n − p + max

{min

{r,

⌈p − t

2

⌉},min

{p − t,

⌈r

2

⌉}}

= max

{min

{n − p + r,n − t −

⌊p − t

2

⌋},min

{n − t,n − p + r −

⌊r

2

⌋}},

which is enough. �Corollary 2.4. Let Q and Q ′ be two non-zero monomial irreducible ideals of S with different associated primeideals. Then

sdepthS S/(Q ∩ Q ′) = max

{min

{dim

(S/Q ′),⌈dim(S/Q ) + dim(S/(Q + Q ′))

2

⌉},

min

{dim(S/Q ),

⌈dim(S/Q ′) + dim(S/(Q + Q ′))

2

⌉}}.

For the proof apply Theorem 1.5 and Theorem 2.3.

Corollary 2.5. Let P and P ′ be two different non-zero monomial prime ideals of S, which are not included onein the other. Then

sdepthS S/(

P ∩ P ′) = max

{min

{dim

(S/P ′),⌈dim(S/P ) + dim(S/(P + P ′))

2

⌉},

min

{dim(S/P ),

⌈dim(S/P ′) + dim(S/(P + P ′))

2

⌉}}.

Proof. For the proof apply Corollary 2.4. �Corollary 2.6. Let � be a simplicial complex in n vertices with only two different facets F , F ′ . Then

sdepth K [�] = max

{min

{∣∣F ′∣∣,⌈ |F | + |F ∩ F ′|2

⌉},min

{|F |,

⌈ |F ′| + |F ∩ F ′|2

⌉}}.

3. An illustration

Let S = K [x1, . . . , x6], Q = (x21, x2

2, x23, x2

4, x1x2x4, x1x3x4), Q ′ = (x24, x5, x6). By our Theorem 2.3 we

get

sdepth S/(

Q ∩ Q ′) � max

{min

{3,

⌈2

2

⌉},min

{2,

⌈3

2

⌉}}= max{1,2} = 2.

On the other hand, we claim that I = ((Q : w) ∩ K [x1, x2, x3]) = (x21, x2

2, x23, x1x2, x1x3) for w = x4 and

sdepth I = 1 < 2 = sdepth(Q ∩ K [x1, x2, x3]). Thus our Proposition 1.4 gives

sdepth S/(

Q ∩ Q ′) � max

{min

{3,

⌈2

2

⌉},min

{2,

⌈3

2

⌉,1

}}= 1.

In this section, we will show that sdepth(S/(Q ∩ Q ′)) = 1.First we prove our claim. Suppose that there exists a Stanley decomposition D of I with

sdepth D � 2. Among the Stanley spaces of D we have five important x21 K [Z1], x2

2 K [Z2], x23 K [Z3],


x1x2 K [Z4], x1x3 K [Z5] for some subsets Zi ⊂ {x1, x2, x3} with |Zi | � 2. If Z4 = {x1, x2, x3} and Z5 con-tains x2 then the last two Stanley spaces will have a non-zero intersection and if Z1 contains x2 thenthe first and the fourth Stanley space will have non-zero intersection. Now if x2 /∈ Z5 and x2 /∈ Z1 thenthe first and the last space will intersect. Suppose that Z4 = {x1, x2}. Then x2 /∈ Z1 (resp. x1 /∈ Z2) be-cause otherwise the intersection of x1x2 K [Z4] with the first Stanley space (resp. the second one) willbe again non-zero. As |Z1|, |Z2| � 2 we get Z1 = {x1, x3}, Z2 = {x2, x3}. But x1 /∈ Z3 because otherwisethe first and the third Stanley space will contain x2

1x23, which is impossible. Similarly, x2 /∈ Z3, which

contradicts |Z3| � 2. The case Z5 = {x1, x3} gives a similar contradiction.Now suppose that Z4 = {x1, x3}. If Z5 ⊃ {x1, x2} we see that the intersection of the last two Stanley

spaces from the above five, contains x21x2x3 and if Z5 = {x2, x3} we see that the intersection of the

same Stanley spaces contains x1x2x3. Contradiction (we saw that Z5 �= {x1, x3})! Hence sdepth D � 1and so sdepth I = 1 using [5].

Next we show that sdepth S/(Q ∩ Q ′) = 1. Suppose that D′ is a Stanley decomposition ofS/(Q ∩ Q ′) such that sdepth S/(Q ∩ Q ′) = 2. We claim that D′ has the form

S/(

Q ∩ Q ′) =(⊕

v K [x5, x6])

⊕(

s⊕i=1

ui K [Zi])

for some monomials v ∈ (K [x1, . . . , x4] \ Q ), ui ∈ (Q ∩ K [x1, . . . , x4]) and Zi ⊂ {x1, x2, x3}. Indeed, letv ∈ (K [x1, . . . , x4] \ Q ). Then vx5, vx6 belong to some Stanley spaces of D′ , let us say uK [Z ], u′K [Z ′].The presence of x5 in u or Z implies that Z does not contain any xi , 1 � i � 3, otherwise uK [Z ] willbe not free over K [Z ]. Thus Z ⊂ {x5, x6}. As |Z | � 2 we get Z = {x5, x6} and similarly Z ′ = {x5, x6}.Thus vx5x6 ∈ (uK [Z ] ∩ u′K [Z ′]) and it follows that u = u′ , Z = Z ′ because the sum in D′ is direct. Itfollows that u|vx5, u|vx6 and so u|v , that is v = u f , f being a monomial in x5, x6. As v ∈ K [x1, . . . , x4]we get f = 1 and so u = v.

A monomial w ∈ (Q \ Q ′) is not a multiple of x5, x6, because otherwise w ∈ Q ′ . Suppose wbelongs to a Stanley space uK [Z ] of D′ . If u ∈ (K [x1, . . . , x4] \ Q ) then as above D′ has also aStanley space uK [x5, x6] and both spaces contains u. This is false since the sum is direct. Thusu ∈ (Q ∩ K [x1, . . . , x4]), which shows our claim.

Hence D′ induces two Stanley decompositions S/Q = ⊕v∈(K [x1,...,x4]\Q ) v K [x5, x6], Q /(Q ∩ Q ′) =⊕s

i=1 ui K [Zi], where ui ∈ (Q ∩ K [x1, . . . , x4]) and Zi ⊂ {x1, x2, x3}. Then we get the following Stanleydecompositions

Q ∩ K [x1, . . . , x3] =s⊕

i=1, ui /∈(x4)

ui K [Zi], I =s⊕

i=1, x4|ui

(ui/x4)K [Zi].

As 2 � mini |Zi | we get sdepth I � 2. Contradiction!

4. A lower bound for Stanley’s depth of some ideals

Let Q , Q ′ be two non-zero irreducible monomial ideals of S such that√

Q = (x1, . . . , xt),√

Q ′ =(xr+1, . . . , xp) for some integers r, t, p with 1 � r � t < p � n, or 0 = r < t < p � n, or 1 � r � t =p � n.

Lemma 4.1. Suppose that p = n, t = r. Then

sdepth(

Q ∩ Q ′) �⌈

r

2

⌉+

⌈n − r

2

⌉� n/2.

Proof. It follows 1 � r < p. Let f ∈ Q ∩ K [x1, . . . , xr], g ∈ Q ′ ∩ K [xr+1, . . . , xn] and M(T ) be the mono-mials from an ideal T . The correspondence ( f , g) → f g defines a map ϕ : M(Q ∩ K [x1, . . . , xr]) ×


M(Q ′ ∩ K [xr+1, . . . , xn]) → M(Q ∩ Q ′), which is injective. If w is a monomial of Q ∩ Q ′ , let ussay w = f g for some monomials f ∈ K [x1, . . . , xr], g ∈ K [xr+1, . . . , xn] then f g ∈ Q and so f ∈ Qbecause the variables xi , i > r are regular on S/Q . Similarly, g ∈ Q ′ and so w = ϕ(( f , g)), that is ϕis surjective. Let D be a Stanley decomposition of Q ∩ K [x1, . . . , xr],

D: Q ∩ K [x1, . . . , xr] =s⊕

i=1

ui K [Zi]

with sdepth D = sdepth(Q ∩ K [x1, . . . , xr]) and D′ a Stanley decomposition of Q ′ ∩ K [xr+1, . . . , xn],

D′: Q ′ ∩ K [xr+1, . . . , xn] =e⊕

j=1

v j K [T j]

with sdepth D′ = sdepth(Q ′ ∩ K [xr+1, . . . , xn]). They induce a Stanley decomposition

D′′: Q ∩ Q ′ =e⊕

j=1

s⊕i=1

ui v j K [Zi ∪ T j]

because of the bijection ϕ . Thus

sdepth(

Q ∩ Q ′) � sdepth D′′ = mini, j

(|Zi| + |T j|)� min

i|Zi| + min

j|T j|

= sdepth D + sdepth D′

= sdepth(

Q ∩ K [x1, . . . , xr]) + sdepth

(Q ′ ∩ K [xr+1, . . . , xn])

=(

r −⌊

r

2

⌋)+

(n − r −

⌊n − r

2

⌋)=

⌈r

2

⌉+

⌈n − r

2

⌉� n/2. �

Remark 4.2. Suppose that n = 8, r = 1. Then by the above lemma we get sdepth(Q ∩ Q ′) �� 1

2 + � 72 = 5. Since |G(Q ∩ Q ′)| = 7 we get by [10,11] the same lower bound sdepth(Q ∩ Q ′) �

8 − � 72 � = 5. If n = 8, r = 2 then by [10,11] we have sdepth(Q ∩ Q ′) � 8 − � 12

2 � = 2 but our previouslemma gives sdepth(Q ∩ Q ′) � � 2

2 + � 62 = 4.

Lemma 4.3. Suppose that p = n. Then

sdepth(

Q ∩ Q ′) �⌈

r

2

⌉+

⌈n − t

2

⌉.

Proof. We show that

Q ∩ Q ′ = (Q ∩ Q ′ ∩ K [xr+1, . . . , xt]

)S

⊕(⊕

w

w(((

Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn]

)),

where w runs in the monomials of K [xr+1, . . . , xt] \ (Q ∩ Q ′). Indeed, a monomial h of S has theform h = f g for some monomials f ∈ K [xr+1, . . . , xt], g ∈ K [x1, . . . , xr, xt+1, . . . , xn]. Since Q , Q ′ are


irreducible we see that h ∈ Q ∩ Q ′ either when f is a multiple of a minimal generator of Q ∩ Q ′ ∩K [xr+1, . . . , xt], or f /∈ (Q ∩ Q ′ ∩ K [xr+1, . . . , xt]) and then

h ∈ f(((

Q ∩ Q ′) : f) ∩ K [x1, . . . , xr, xt+1, . . . , xn]).

Let D be a Stanley decomposition of (Q ∩ Q ′ ∩ K [xr+1, . . . , xt])S ,

D: (Q ∩ Q ′ ∩ K [xr+1, . . . , xt]

)S =

s⊕i=1

ui K [Zi]

with sdepth D = sdepth(Q ∩ Q ′ ∩ K [xr+1, . . . , xt])S and for all w ∈ (K [xr+1, . . . , xt] \ (Q ∩ Q ′)), letD w be a Stanley decomposition of ((Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn],

D w : ((Q ∩ Q ′) : w

) ∩ K [x1, . . . , xr, xt+1, . . . , xn] =⊕

w

⊕j

v w j K [T w j]

with sdepth D w = sdepth(((Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn]). Since K [xr+1, . . . , xt] \ (Q ∩ Q ′)contains just a finite set of monomials we get a Stanley decomposition of Q ∩ Q ′ ,

D′: Q ∩ Q ′ =(

s⊕i=1

ui K [Zi])

⊕(⊕

w

⊕j

w v w j K [T w j])

,

where w runs in the monomials of K [xr+1, . . . , xt] \ (Q ∩ Q ′). Then

sdepth D′ = minw

{sdepth D, sdepth D w}= min

w

{sdepth

(Q ∩ Q ′ ∩ K [xr+1, . . . , xt]

)S,

sdepth(((

Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn]

)}.

But ((Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn] is still an intersection of two irreducible ideals and

sdepth(((

Q ∩ Q ′) : w) ∩ K [x1, . . . , xr, xt+1, . . . , xn]) �

⌈r

2

⌉+

⌈n − t

2

⌉

by Lemma 4.1. We have sdepth(Q ∩ Q ′ ∩ K [xr+1, . . . , xt]) � 1 and so

sdepth(

Q ∩ Q ′ ∩ K [xr+1, . . . , xt])

S � 1 + n − t + r

by [9, Lemma 3.6]. Thus

sdepth(

Q ∩ Q ′) � sdepth D′ �⌈

r

2

⌉+

⌈n − t

2

⌉.

Note that the proof goes even when 0 � r < t � n (anyway sdepth Q ∩ Q ′ � 1 if n = t , r = 0). �Lemma 4.4.

sdepth(

Q ∩ Q ′) � n − p +⌈

r

2

⌉+

⌈p − t

2

⌉.


Proof. As usual we see that there are now (n − p) free variables and it is enough to apply [9, Lemma3.6] and Lemma 4.3. �Theorem 4.5. Let Q and Q ′ be two non-zero irreducible monomial ideals of S. Then

sdepthS

(Q ∩ Q ′) � dim

(S/

(Q + Q ′)) +

⌈dim(S/Q ′) − dim(S/(Q + Q ′))

2

⌉

+⌈

dim(S/Q ) − dim(S/(Q + Q ′))2

⌉

�⌈

dim(S/Q ′) + dim(S/Q )

2

⌉.

Proof. After renumbering of variables, we may suppose as above that√

Q = (x1, . . . , xt),√

Q ′ =(xr+1, . . . , xp) for some integers r, t, p with 1 � r � t < p � n, or 0 = r < t < p � n, or 1 � r �t = p � n. If n = p, r = 0 then

√Q ⊂ √

Q ′ and the inequality is trivial. It is enough to ap-ply Lemma 4.4 because n − p = dim(S/(Q + Q ′)), r = dim(S/Q ′) − dim(S/(Q + Q ′)), p − t =dim(S/Q ) − dim(S/(Q + Q ′)). �Remark 4.6. If Q , Q ′ are non-zero irreducible monomial ideals of S with

√Q = √

Q ′ then we havesdepthS (Q ∩ Q ′) � 1 + dim S/Q .

Example 4.7. Let S = K [x1, x2], Q = (x1), Q ′ = (x21, x2). We have

sdepth(

Q ∩ Q ′) �⌈


2

⌉=

⌈1 + 0

2

⌉= 1

by the above theorem. As Q ∩ Q ′ is not a principle ideal its Stanley depth is < 2. Thus

sdepth(

Q ∩ Q ′) = 1.

Example 4.8. Let S = K [x1, x2, x3, x4, x5], Q = (x1, x2, x23), Q ′ = (x3, x4, x5). As dim(S/(Q + Q ′)) = 0,

dim S/Q = 2 and dim S/Q ′ = 2 we get

sdepth(

Q ∩ Q ′) �⌈


2

⌉=

⌈2 + 2

2

⌉= 2

by the above theorem. Note also that

sdepth(

Q ∩ Q ′ ∩ K [x1, x2, x4, x5]) = sdepth(x1x4, x1x5, x2x4, x2x5)K [x1, x2, x4, x5] = 3,

and

sdepth(((

Q ∩ Q ′) : x3) ∩ K [x1, x2, x4, x5]

) = sdepth((x1, x2)K [x1, x2, x4, x5]

)= 4 −

⌊2

2

⌋= 3,

by [15]. But sdepth(Q ∩ Q ′) � 3 because of the following Stanley decomposition


Q ∩ Q ′ = x1x4 K [x1, x4, x5] ⊕ x1x5 K [x1, x2, x5] ⊕ x2x4 K [x1, x2, x4] ⊕ x2x5 K [x2, x4, x5]⊕ x2

3 K [x3, x4, x5] ⊕ x2x3 K [x2, x3, x4] ⊕ x1x3 K [x1, x2, x3] ⊕ x1x3x4 K [x1, x2, x4, x5]⊕ x1x3x5 K [x1, x3, x5] ⊕ x2x3x5 K [x2, x3, x4, x5] ⊕ x1x2x4x5 K [x1, x2, x4, x5]⊕ x1x2

3x4 K [x1, x3, x4, x5] ⊕ x1x2x3x5 K [x1, x2, x3, x5] ⊕ x1x2x23x4 K [x1, x2, x3, x4, x5].

5. Applications

Let I ⊂ S be a non-zero monomial ideal. A. Rauf presented in [14] the following:

Question 5.1. Does it hold the inequality

sdepth I � 1 + sdepth S/I?

The importance of this question is given by the following:

Proposition 5.2. Suppose that Stanley’s Conjecture holds for cyclic S-modules and the above question has apositive answer for all monomial ideals of S. Then Stanley’s Conjecture holds for all monomial ideals of S.

For the proof note that sdepth I � 1 + sdepth S/I � 1 + depth S/I = depth I .

Remark 5.3. In [12] it is proved that Stanley’s Conjecture holds for all multigraded cycle modules overS = K [x1, . . . , x5]. If the above question has a positive answer then Stanley’s Conjecture holds for allmonomial ideals of S . Actually this is true for all square free monomial ideals of S as [13] shows.

We show that the above question holds for the intersection of two non-zero irreducible monomialideals.

Proposition 5.4. Question 5.1 has a positive answer for intersections of two non-zero irreducible monomialideals.

Proof. First suppose that Q , Q ′ have different associated prime ideals. After renumbering of variableswe may suppose as above that

√Q = (x1, . . . , xt),

√Q ′ = (xr+1, . . . , xp) for some integers r, t, p with

1 � r � t < p � n, or 0 = r < t < p � n, or 1 � r � t = p � n. Then

sdepth(

Q ∩ Q ′) � n − p +⌈

r

2

⌉+

⌈p − t

2

⌉

by Lemma 4.4. Note that

sdepth(

S/(

Q ∩ Q ′)) = n − p + max

{min

{r,

⌈p − t

2

⌉},min

{p − t,

⌈r

2

⌉}}

by Corollary 2.4. Thus

1 + sdepth(

S/(

Q ∩ Q ′)) � n − p +⌈

r

2

⌉+

⌈p − t

2

⌉� sdepth

(Q ∩ Q ′).

Finally, if Q , Q ′ have the same associated prime ideal then sdepth(Q ∩ Q ′) � 1 + dim S/Q by Re-mark 4.6 and so sdepth(Q ∩ Q ′) � 1 + sdepth S/(Q ∩ Q ′). �


Next we will show that Stanley’s Conjecture holds for intersections of two primary monomialideals. We start with a simple lemma.

Lemma 5.5. Let Q , Q ′ be two primary ideals in S = K [x1, . . . , xn]. Suppose√

Q = (x1, . . . , xt) and√

Q =(xr+1, . . . , xp) for integers 0 � r � t � p � n. Then sdepth(S/(Q ∩ Q ′)) � depth(S/(Q ∩ Q ′)), that is Stan-ley’s Conjecture holds for S/(Q ∩ Q ′).

Proof. If either r = 0, or t = p then depth S/(Q ∩ Q ′) � n− p � sdepth(S/(Q ∩ Q ′)) by [9, Lemma 3.6].Now suppose that r > 0, t < p and let S ′ = K [x1, . . . , xp] and q = Q ∩ S ′ , q′ = Q ′ ∩ S ′ . Consider thefollowing exact sequence of S ′-modules

0 → S ′/(q ∩ q′) → S ′/q ⊕ S ′/q′ → S ′/

(q + q′) → 0.

By Lemma 1.2

depth(

S ′/q ⊕ S ′/q′) = min{

depth(

S ′/q),depth

(S ′/q′)}

= min{

dim(

S ′/q),dim

(S ′/q′)}

= min{r, p − t} � 1 > 0

= depth(

S ′/(q + q′)).

Thus by Depth Lemma (see e.g. [4])

depth(

S ′/q ∩ q′) = depth(

S ′/(q + q′)) + 1 = 1.

But sdepth(S ′/(q ∩ q′)) � 1 by [5, Corollary 1.6] and so

sdepth(

S/(

Q ∩ Q ′)) = sdepth(

S ′/(q ∩ q′)) + n − p � 1 + n − p

= n − p + depth(

S ′/(q ∩ q′))

= depth(

S/(

Q ∩ Q ′))by [9, Lemma 3.6]. �Theorem 5.6. Let Q , Q ′ be two non-zero irreducible ideals of S. Then sdepth(Q ∩ Q ′) � depth(Q ∩ Q ′),that is Stanley’s Conjecture holds for Q ∩ Q ′ .

Proof. By Proposition 5.4, Question 5.1 has a positive answer, so by the proof of Proposition 5.2 it isenough to know that Stanley’s Conjecture holds for S/(Q ∩ Q ′). This is given by the above lemma. �

Next we consider the cycle module given by an irredundant intersection of 3 irreducible ideals.

Lemma 5.7. Let Q 1 , Q 2 , Q 3 be three non-zero irreducible monomial ideals of S = K [x1, . . . , xn]. Then

sdepth((Q 2 ∩ Q 3)/(Q 1 ∩ Q 2 ∩ Q 3)

)� dim

(S/(Q 1 + Q 2 + Q 3)

) +⌈

dim(S/(Q 1 + Q 2)) − dim(S/(Q 1 + Q 2 + Q 3))

2

⌉


+⌈

dim(S/(Q 1 + Q 3)) − dim(S/(Q 1 + Q 2 + Q 3))

2

⌉

�⌈

dim(S/(Q 1 + Q 2)) + dim(S/(Q 1 + Q 3))

2

⌉.

If Q 3 ⊂ Q 1 + Q 2 then

sdepth((Q 2 ∩ Q 3)/(Q 1 ∩ Q 2 ∩ Q 3)

)�

⌈dim(S/Q 1) + dim(S/(Q 1 + Q 3))

2

⌉.

Proof. Renumbering the variables we may assume that√

Q 1 = (x1, . . . , xt) and√

Q 2 + Q 3 =(xr+1, . . . , xp), where 0 � r � t < p � n. If t = p then

√Q 1 + Q 2 = √

Q 1 + Q 3 and the inequalityis trivial by [9, Lemma 3.6]. Let S ′ = K [x1, . . . , xp] and q1 = Q 1 ∩ S ′ , q2 = Q 2 ∩ S ′ , q3 = Q 3 ∩ S ′. Wehave a canonical injective map (q2 ∩ q3)/(q1 ∩ q2 ∩ q3) → S ′/q1. Now by Lemma 1.1, we have

S ′/q1 =⊕

uK [xt+1, . . . , xp]

and so

(q2 ∩ q3)/(q1 ∩ q2 ∩ q3) =⊕(

(q2 ∩ q3) ∩ uK [xt+1, . . . , xp]),where u runs in the monomials of K [x1, . . . , xt] \ (q1 ∩ K [x1, . . . , xt]). If u ∈ K [x1, . . . , xr] then

(q2 ∩ q3) ∩ uK [xt+1, . . . , xp] = u(q2 ∩ q3 ∩ K [xt+1, . . . , xp])

and if u /∈ K [x1, . . . , xr] then

(q2 ∩ q3) ∩ uK [xt+1, . . . , xp] = u((

(q2 ∩ q3) : u) ∩ K [xt+1, . . . , xp]).

Since (q2 ∩ q3) : u is still an intersection of irreducible monomial ideals we get by Lemma 4.3 that

sdepth((

(q2 ∩ q3) : u) ∩ K [xt+1, . . . , xp])

�⌈

dim K [xt+1, . . . , xp]/q2 ∩ K [xt+1, . . . , xp]2

⌉+

⌈dim K [xt+1, . . . , xp]/q3 ∩ K [xt+1, . . . , xp]

2

⌉.

Also we have

q2/(q1 ∩ q2) =⊕

u(q2 ∩ K [xt+1, . . . , xp]),

and it follows

S ′/(q1 + q2) ∼= (S ′/q1

)/(q2/(q1 ∩ q2)

) =⊕

u(

K [xt+1, . . . , xp]/q2 ∩ K [xt+1, . . . , xp]).Thus dim S ′/(q1 + q2) = dim K [xt+1, . . . , xp]/q2 ∩ K [xt+1, . . . , xp] and similarly

dim S ′/(q1 + q3) = dim K [xt+1, . . . , xp]/q3 ∩ K [xt+1, . . . , xp].Hence


sdepth((q2 ∩ q3)/(q1 ∩ q2 ∩ q3)

)�

⌈dim(S ′/(q1 + q2))

2

⌉+

⌈dim(S ′/(q1 + q3))

2

⌉

=⌈

dim(S/(Q 1 + Q 2)) − dim(S/(Q 1 + Q 2 + Q 3))

2

⌉

+⌈

dim(S/(Q 1 + Q 3)) − dim(S/(Q 1 + Q 2 + Q 3))

2

⌉.

If Q 3 ⊂ Q 1 + Q 2 then (q2 ∩ q3) ∩ K [xt+1, . . . , xp] = q3 ∩ K [xt+1, . . . , xp] and so

sdepthS ′(q2 ∩ q3)/(q1 ∩ q2 ∩ q3) � sdepth((q2 ∩ q3) ∩ K [xt+1, . . . , xp])

= p − t −⌊

ht(q3 ∩ K [xt+1, . . . , xp])2

⌋

=⌈

p − t + dim K [xt+1, . . . , xp]/q3 ∩ K [xt+1, . . . , xp]2

⌉

=⌈

dim(S ′/q1) + dim(S ′/(q1 + q3))

2

⌉.

Now it is enough to apply [9, Lemma 3.6]. �Proposition 5.8. Let Q 1 , Q 2 , Q 3 be three non-zero irreducible ideals of S and R = S/Q 1 ∩ Q 2 ∩ Q 3 . Supposethat dim S/(Q 1 + Q 2 + Q 3) = 0. Then

sdepth R � max

{min

{sdepth S/(Q 2 ∩ Q 3),

⌈dim(S/(Q 1 + Q 2))

2

⌉+

⌈dim(S/(Q 1 + Q 3))

2

⌉},

min


⌈dim(S/(Q 1 + Q 2))

2

⌉+

⌈dim(S/(Q 2 + Q 3))

2

⌉},

min


⌈dim(S/(Q 3 + Q 2))

2

⌉+

⌈dim(S/(Q 1 + Q 3))

2

⌉}}.

For the proof apply Lemma 1.3 and Lemma 5.7.

Theorem 5.9. Let Q 1 , Q 2 , Q 3 be three non-zero irreducible ideals of S and R = S/(Q 1 ∩ Q 2 ∩ Q 3). Thensdepth R � depth R, that is Stanley’s Conjecture holds for R.

Proof. Applying [9, Lemma 3.6] we may reduce the problem to the case when

dim S/(Q 1 + Q 2 + Q 3) = 0.

If one of the Q i has dimension 0 then depth R = 0 and there exists nothing to show. Assume thatall Q i have dimension > 0. If one of the Q i has dimension 1 then depth R = 1 and by [5] (or [7]) weget sdepth R � 1 = depth R . From now on we assume that all Q i have dimension > 1.

If Q 1 + Q 2 has dimension 0 and Q 3 �⊂ Q 1 + Q 2 then from the exact sequence

0 → R → S/Q 1 ⊕ S/Q 2 ∩ Q 3 → S/(Q 1 + Q 2) ∩ (Q 1 + Q 3) → 0,

we get depth R = 1 by Depth Lemma and we may apply [5] (or [7]) to get as above sdepth R � 1 =depth R . If Q 3 ⊂ Q 1 + Q 2 then by Lemma 1.3, Theorem 5.6 and Lemma 5.7 we have


sdepth R � min

{depth S/(Q 2 ∩ Q 3),

⌈dim(S/Q 1) + dim(S/(Q 1 + Q 3))

2

⌉}

� 1 + min{

dim S/(Q 2 + Q 3),dim S/(Q 1 + Q 3)}

= depth R

from the above exact sequence and a similar one. Thus we may suppose that Q 1 + Q 2, Q 2 + Q 3,Q 1 + Q 3 have dimension � 1. Then from the exact sequence

0 → S/(Q 1 + Q 2) ∩ (Q 1 + Q 3) → S/(Q 1 + Q 2) ⊕ S/(Q 1 + Q 3) → S/(Q 1 + Q 2 + Q 3) → 0

we get by Depth Lemma depth S/(Q 1 + Q 2) ∩ (Q 1 + Q 3) = 1. Renumbering Q i we may suppose thatdim(Q 2 + Q 3) � max{dim(Q 1 + Q 3),dim(Q 2 + Q 1)}. Using Proposition 5.8 we have

sdepth R � min

{sdepth S/Q 2 ∩ Q 3,

⌈dim(S/(Q 1 + Q 2))

2

⌉+

⌈dim(S/(Q 1 + Q 3))

2

⌉}.

We may suppose that sdepth R < dim S/Q i because otherwise sdepth R � dim S/Q i � depth R . Thususing Theorem 1.5 we get

sdepth R � min

{⌈dim S/Q 3 + dim S/(Q 2 + Q 3)

2

⌉,

⌈dim(S/(Q 1 + Q 2))

2

⌉+

⌈dim(S/(Q 1 + Q 3))

2

⌉}.

If Q 1 �⊂ √Q 3 then dim S/Q 3 > dim S/(Q 1 + Q 3) and we get

dim S/Q 3 + dim S/(Q 2 + Q 3) > dim(

S/(Q 1 + Q 2)) + dim

(S/(Q 1 + Q 3)

)because dim S/(Q 2 + Q 3) is maxim by our choice. It follows that

sdepth R �⌈

dim(S/(Q 1 + Q 2))

2

⌉+

⌈dim(S/(Q 1 + Q 3))

2

⌉� 2.

But from the first above exact sequence we get depth R = 2 with Depth Lemma, that is sdepth R �depth R .

If Q 1 �⊂ √Q 2 we note that dim S/Q 2 + dim S/(Q 2 + Q 3) > dim(S/(Q 1 + Q 2))+ dim(S/(Q 1 + Q 3))

and we proceed similarly as above with Q 2 instead Q 3. Note also that if Q 1 ⊂ √Q 2 and Q 1 ⊂√

Q 3 we get dim S/(Q 2 + Q 3) � dim S/(Q 2 + Q 1) = dim S/Q 2, respectively dim S/(Q 2 + Q 3) �dim S/(Q 3 + Q 1) = dim S/Q 3. Thus Q 1 ⊂ √

Q 3 = √Q 2 and it follows sdepth R � dim S/Q 2, which

is a contradiction. �References

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http://dx.doi.org/10.1016/j.jcta.2009.07.008


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computing the stanley depth

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